令2kπ+π/2＜2x+π/4＜2kπ+3π/2(k∈Z)
得2kπ+π/4＜2x＜2kπ+5π/4(k∈Z)
即kπ+π/8＜x＜kπ+5π/8(k∈Z)
所以y=sin(2x+π/4）的单调递减区间是(kπ+π/8,kπ+5π/8)(k∈Z)
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